Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm ${f}'\left( x \right)$ liên tục trên $\left[ -2;1 \right]$. Hình bên là đồ thị của hàm số $y={f}'\left( x \right)$. Đặt $g\left( x \right)=f\left( x \right)-\dfrac{{{x}^{2}}}{2}$. Khẳng định nào sau đây đúng?
A. $g\left( 1 \right)<g\left( -2 \right)<g\left( 0 \right)$
B. $g\left( 0 \right)<g\left( 1 \right)<g\left( -2 \right)$
C. $g\left( -2 \right)<g\left( 1 \right)<g\left( 0 \right)$
D. $g\left( 0 \right)<g\left( -2 \right)<g\left( 1 \right)$
Ta có ${g}'\left( x \right)={f}'\left( x \right)-x$.
$\int\limits_{-2}^{0}{{g}'\left( x \right)d\text{x}}=g\left( 0 \right)-g\left( -2 \right)=\int\limits_{-2}^{0}{\left[ {f}'\left( x \right)-x \right]d\text{x}}>0\Rightarrow g\left( 0 \right)-g\left( -2 \right)>0$.
$\int\limits_{0}^{1}{{g}'\left( x \right)d\text{x}}=\int\limits_{0}^{1}{\left[ {f}'\left( x \right)-x \right]d\text{x}}<0\Rightarrow g\left( 1 \right)-g\left( 0 \right)<0\Rightarrow g\left( 1 \right)<g\left( 0 \right)$
Mặt khác, ta có $g\left( 0 \right)-g\left( 1 \right)=\int\limits_{0}^{1}{\left[ x-{f}'\left( x \right) \right]d\text{x}}>0$.
Từ hình vẽ, ta có $g\left( 0 \right)-g\left( -2 \right)>g\left( 0 \right)-g\left( 1 \right)\Rightarrow g\left( 1 \right)>g\left( -2 \right)$.
Vậy $g\left( -2 \right)<g\left( 1 \right)<g\left( 0 \right)$.
A. $g\left( 1 \right)<g\left( -2 \right)<g\left( 0 \right)$
B. $g\left( 0 \right)<g\left( 1 \right)<g\left( -2 \right)$
C. $g\left( -2 \right)<g\left( 1 \right)<g\left( 0 \right)$
D. $g\left( 0 \right)<g\left( -2 \right)<g\left( 1 \right)$
Ta có ${g}'\left( x \right)={f}'\left( x \right)-x$.
$\int\limits_{-2}^{0}{{g}'\left( x \right)d\text{x}}=g\left( 0 \right)-g\left( -2 \right)=\int\limits_{-2}^{0}{\left[ {f}'\left( x \right)-x \right]d\text{x}}>0\Rightarrow g\left( 0 \right)-g\left( -2 \right)>0$.
$\int\limits_{0}^{1}{{g}'\left( x \right)d\text{x}}=\int\limits_{0}^{1}{\left[ {f}'\left( x \right)-x \right]d\text{x}}<0\Rightarrow g\left( 1 \right)-g\left( 0 \right)<0\Rightarrow g\left( 1 \right)<g\left( 0 \right)$
Mặt khác, ta có $g\left( 0 \right)-g\left( 1 \right)=\int\limits_{0}^{1}{\left[ x-{f}'\left( x \right) \right]d\text{x}}>0$.
Từ hình vẽ, ta có $g\left( 0 \right)-g\left( -2 \right)>g\left( 0 \right)-g\left( 1 \right)\Rightarrow g\left( 1 \right)>g\left( -2 \right)$.
Vậy $g\left( -2 \right)<g\left( 1 \right)<g\left( 0 \right)$.
Đáp án C.